# mod $p$ irreducibility test over $\mathbb{Z}$

I have a question about the following result:

If $$f ∈ \mathbb{Z}[X]$$ is primitive and there is a prime $$p$$ not dividing the leading coefficient of $$f$$ such that $$f$$ is irreducible in $$(\mathbb{Z}/p \mathbb{Z})[X]$$ then $$f(x)$$ is irreducible in $$\mathbb{Z}[X]$$.

My question: Does this result still hold if $$p$$ is composite? I have looked at a few proofs of this statement and I’m not sure how the fact that $$p$$ is prime is being used. Any insight would be appreciated.

• The same question was asked here, and I suppose there are more. Mar 17, 2022 at 17:53
• @DietrichBurde Obviously we require that the leading coefficient is a unit Mar 17, 2022 at 19:23
• Yes, but still the point is that then $\Bbb Z/n$ has zero divisors. Mar 17, 2022 at 19:26
• @DietrichBurde No the point is that we need to restrict to polynomials whose leading coefficient is a unit and that it works well. Mar 17, 2022 at 19:28
• So the result does not hold as the OP wanted, but only with monic polynomials? The other linked question also had this problem. Mar 17, 2022 at 19:31